(a) As we observed in the proof of Theorem REF ,
the restriction of \(f\)  to the subgroup \(B_{2n+1}\)  of \(B_{2n+2}\)  is the monodromy representation considered in [1]}.
The kernel of this restriction is the hyperelliptic Torelli group \({\mathcal {S}}{\mathcal {I}}_n^1\)  considered in [2]}.
In loc. cit. Brendle, Margalit, Putman prove that \({\mathcal {S}}{\mathcal {I}}_n^1\)  is isomorphic to a subgroup \({\mathcal {B}}{\mathcal {I}}_{2n+1}\) 
of the mapping class group of the disk with \(2n+1\)  marked points.
By Theorem C of [2]} and the comments thereafter the subgroup \({\mathcal {B}}{\mathcal {I}}_{2n+1}\) 
is generated by squares of Dehn twists about curves in the disk surrounding exactly 3 or 5 marked points.
Now under the standard identification of the mapping class group with the
braid group \(B_{2n+1}\)  (see e.g. [4]}), the square of a Dehn twist about a curve surrounding 3
(resp. 5) marked points corresponds in the braid group to a conjugate of the element \(\Delta _3^4\)  (resp. of \(\Delta _5^4\) ).
Thus, \(\mathrm {Ker}(\bar{f}) \cap B_{2n+1} = \mathrm {Ker}(\bar{f}) \cap P_{2n+1}\)  is the normal closure of \(\Delta _3^4\)  and \(\Delta _5^4\) .
To determine the whole kernel inside \(P_{2n+2}\) , we use the fact that
\(P_{2n+2}\)  is the semi-direct product of a free group by \(P_{2n+1}\)  (see [4]}).
This free group has \(2n+1\)  generators \(A_{i,2n+2}\)  (\(i= 1, \ldots , 2n+1\) ) defined by \(A_{2n+1,2n+2} = \sigma _{2n+1}^2\) 
and
\(A_{i,2n+2} = (\sigma _{2n+1} \sigma _{2n} \cdots \sigma _{i+1}) \, \sigma _i^2 \,(\sigma _{2n+1} \sigma _{2n} \cdots \sigma _{i+1})^{-1}\) 
when \(1 \le i \le 2n\) . The following expression for \(A_{i,2n+2}\)  will be used in the sequel:
\(A_{i,2n+2} = (\sigma _{2n} \sigma _{2n-1} \cdots \sigma _i)^{-1} \, \sigma _{2n+1}^2 \,(\sigma _{2n} \sigma _{2n-1} \cdots \sigma _i). \quad (1 \le i \le 2n)\) 
To derive (REF ) from (REF ) use the braid relations or draw a picture.
Recall the special element \(\alpha _n = \gamma _n \beta _n \gamma _n^{-1} \beta _n\)  given by (REF ).
The element \(\gamma _n\)  belongs to \(P_{2n}\)  and \(\beta _n\)  to \(B_{2n}\sigma _{2n+1}^{\pm 1}\) .
Since \(\sigma _{2n+1}\)  commutes with \(B_{2n}\)  we have
\(\alpha _n\in B_{2n}\sigma _{2n+1}^{\pm 2}\cap P_{2n+2}=P_{2n}\sigma _{2n+1}^{\pm 2}\) .
Consequently, \(A_{2n+1,2n+2} = \sigma _{2n+1}^2\)  belongs to \(P_{2n} \alpha _n^{\pm 1}\) .
Let us deal with \(A_{i,2n+2}\)  when \(1 \le i \le 2n\) .
Since \(\sigma _{2n} \sigma _{2n-1} \cdots \sigma _i\)  belongs to \(B_{2n+1}\) , we see from (REF )
and what we established for \(\sigma _{2n+1}^2\)  that \(A_{i,2n+2}\)  is conjugate under \(B_{2n+1}\)  of
an element of \(P_{2n} \alpha _n\)  or of \(P_{2n}\alpha _n^{-1}\) .
So, at the cost of adding \(\alpha _n\) , we have reduced \(\mathrm {Ker}(\bar{f})\)  to its intersection with \(P_{2n+1}\) .
In view of the above considerations, this completes the proof of Part (a).
(b) Let \(N\)  be the normal closure of \([\sigma _3, \Delta _3^4]\) , \(\Delta _5^4 \, \Delta _3^{-16}\)  and \(\alpha _n\) .
We have \(N \subset \mathrm {Ker}(f)\) : this follows from the centrality of \(f(\Delta _3^4) = w_1^4\) ,
from Corollary REF  (a) applied to \(i=2\) , and from Proposition REF .
Let us next remark that \(N\)  contains the commutator \([\beta , \Delta _3^4]\)  for each braid \(\beta \) .
Indeed, since \(\Delta _3^4\)  is a product of \(\sigma _1\)  and \(\sigma _2\) , it commutes with all generators \(\sigma _i\) 
and their inverses with \(4 \le i \le 2n+1\) .
On the other hand \(\Delta _3^4\)  is central in \(B_3\) ; therefore it commutes with \(\sigma _1\)  and \(\sigma _2\)  and their inverses.
Since by definition \(N\)  contains \([\sigma _3, \Delta _3^4]\)  and we have
\([\sigma _3^{-1}, \Delta _3^4] = \sigma _3^{-1} [\sigma _3, \Delta _3^4]^{-1} \sigma _3\) ,
it contains all commutators of the form \([\sigma _i^{\pm 1}, \Delta _3^4]\) . Using the commutator identities
\([\beta _1 \beta _2, \Delta _3^4] = \beta _1 [\beta _2, \Delta _3^4] \beta _1^{-1} \, [\beta _1, \Delta _3^4]\) 
for \(\beta _1, \beta _2 \in B_{2n+2}\) ,
we conclude by induction on the length of expression of a braid in the generators \(\sigma _i^{\pm 1}\) .
The kernels of \(f\)  and of \(\bar{f}\)  are connected by the short exact sequence
\(1 \rightarrow \mathrm {Ker}(f) \longrightarrow \mathrm {Ker}(\bar{f}) \overset{f}{\longrightarrow }\langle w_1^4 \rangle \rightarrow 1.\) 
Indeed, clearly \(\mathrm {Ker}(f)\)  sits inside \(\mathrm {Ker}(\bar{f})\)  as a normal subgroup.
If \(\bar{f}(\beta )= 1\) , then \(f(\beta )\)  belongs to the kernel of \(\pi : \operatorname{St}(C_n,{\mathbb {Z}}) \rightarrow \operatorname{Sp}_{2n}({\mathbb {Z}})\) ,
which we know to be infinite cyclic generated by \(w_1^4\) .
The homomorphism \(f: \mathrm {Ker}(\bar{f}) \rightarrow \langle w_1^4 \rangle \)  is surjective since
\(w_1^4 = f(\Delta _3^4)\)  and \(\bar{f}(\Delta _3^4) = 1\) .
Moreover, since \(\langle w_1^4 \rangle \cong {\mathbb {Z}}\) , the short exact sequence (REF ) is split
with splitting \(\langle w_1^4 \rangle \rightarrow \mathrm {Ker}(\bar{f})\)  given by \(w_1^4 \mapsto \Delta _3^4\) .
The short exact sequence (REF ) induces the quotient short exact sequence
\(1 \rightarrow \mathrm {Ker}(f)/N \longrightarrow \mathrm {Ker}(\bar{f})/N \overset{f}{\longrightarrow }\langle w_1^4 \rangle \rightarrow 1.\) 
To conclude that \(\mathrm {Ker}(f) = N\) , it suffices to check that
\(f: \mathrm {Ker}(\bar{f})/N \rightarrow \langle w_1^4 \rangle \)  is injective.
Now, by Part (a) each element of \(\mathrm {Ker}(\bar{f})\)  is a product of conjugates of the braids \(\Delta _3^4\) , \(\Delta _5^4\) ,
\(\alpha _n\)  and their inverses. Since \(\alpha _n \equiv 1\)  and \(\Delta _5^4 \equiv (\Delta _3^4)^4\)  modulo \(N\) ,
each element of \(\mathrm {Ker}(\bar{f})\)  is equal modulo \(N\)  to a product of conjugates of \(\Delta _3^4\)  and of its inverse.
Since \([\beta , \Delta _3^4] \in N\)  for any braid \(\beta \) , we have \(\beta \Delta _3^4 \beta ^{-1} \equiv \Delta _3^4\)  modulo \(N\) .
Hence, every element of \(\mathrm {Ker}(\bar{f})\)  is equal modulo \(N\)  to a power of \(\Delta _3^4\) .
In other words, \(\mathrm {Ker}(\bar{f})/N = N \langle \Delta _3^4 \rangle /N\) . Now, since \(w_1^4\)  is of infinite order, \(f\)  sends
\(N \langle \Delta _3^4 \rangle /N\)  injectively into \(\langle w_1^4 \rangle \) .
We conclude that \(f: \mathrm {Ker}(\bar{f})/N = N \langle \Delta _3^4 \rangle /N \rightarrow \langle w_1^4 \rangle \)  is injective.
Therefore, \(\mathrm {Ker}(f) = N\) .
